JEE Main 2023 — Parabola Question with Solution

Given,

The $x$-intercept of a focal chord of the parabola $y^2=8x+4y+4$ is $3$,

Now simplifying equation of the given parabola we get,

$y^2=8x+4y+4$

$\Rightarrow {\left(y-2\right)}^2=8\left(x+1\right)$

Now on comparing with standard parabola $y^2=4ax$ we get,

$a=2, X=x+1, Y=y-2$

Hence, the vertex is $\left(-1,2\right)$ and focus is $\left(1,2\right)$

So, equation of line passing through focus is given by $y-2=m\left(x-1\right)$

Now, putting $\left(3,0\right)$ {as given $x-$intercept of focal chord is $3$} in the above line we get, $m=-1$ 

So, equation of chord will be $y=3-x$,

Now putting the value of $y=3-x$ in parabola ${\left(y-2\right)}^2=8\left(x+1\right)$ we get,

$\Rightarrow {\left(1-x\right)}^2=8\left(x+1\right)$

$\Rightarrow x^2-10x-7=0$

$\Rightarrow \left|x_1-x_2\right|=\sqrt{{10}^2+4\times 7}=\sqrt{128}$

$\Rightarrow \left|y_1-y_2\right|=\left|3-x_1-\left(3-x_2\right)\right|=\left|x_2-x_1\right|=\sqrt{128}$

Length of focal chord $\sqrt{{\left|x_2-x_1\right|}^2+{\left|y_2-y_1\right|}^2}=\sqrt{256}=16$.